Problem: Is ${661149}$ divisible by $3$ ?
Solution: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {661149}= &&{6}\cdot100000+ \\&&{6}\cdot10000+ \\&&{1}\cdot1000+ \\&&{1}\cdot100+ \\&&{4}\cdot10+ \\&&{9}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {661149}= &&{6}(99999+1)+ \\&&{6}(9999+1)+ \\&&{1}(999+1)+ \\&&{1}(99+1)+ \\&&{4}(9+1)+ \\&&{9} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {661149}= &&\gray{6\cdot99999}+ \\&&\gray{6\cdot9999}+ \\&&\gray{1\cdot999}+ \\&&\gray{1\cdot99}+ \\&&\gray{4\cdot9}+ \\&& {6}+{6}+{1}+{1}+{4}+{9} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${661149}$ is divisible by $3$ if ${ 6}+{6}+{1}+{1}+{4}+{9}$ is divisible by $3$ Add the digits of ${661149}$ $ {6}+{6}+{1}+{1}+{4}+{9} = {27} $ If ${27}$ is divisible by $3$ , then ${661149}$ must also be divisible by $3$ ${27}$ is divisible by $3$, therefore ${661149}$ must also be divisible by $3$.